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\begin{document}

\preprint{ADP-11-45/T767}

\title{Limits on the Observable Dynamics of Mixed States}
 
\pacs{03.65.Ta, 03.65.Ud, 03.67.-a}

\author{Cael L. Hasse}
\email{Electronic address: cael.hasse@adelaide.edu.au}
\affiliation{Special Research Centre for the Subatomic Structure of Matter and Department of Physics, University of Adelaide 5005, Australia.}

\begin{abstract}
It is shown that the observability of a large class of operations on mixed states is fundamentally limited. This is done by first showing the existence of symmetries of completely mixed states which extend the symmetry of envariance - environment-induced invariance - to trace-preserving transformations where the operation elements $E_j$ in an operator sum representation also satisfy $\sum_j E_j E^{\dagger}_j = \mathbb{I}$. This class includes unitary and perfect premeasurement operations. For less than maximum entropy mixed states, an upper bound on the trace distance between an untransformed state and a state transformed by one of these operations is derived. The bound involves only a new measure of the level of mixedness of a state, which we introduce.
\end{abstract}

\maketitle

\section{Introduction}

Given a particular quantum state that is subjected to a class of operations on the state, does the state change? And if so, how observable is that change? This question is related to the study of the preservation of information subject to quantum processes \cite{Chuang:1997, Nielsen:1996, Gilchrist:2005} by Nielsen {\it et al}. Their work focuses on the slightly different question of quantifying changes to quantum states given specific operations. By asking our initial question instead of that raised by Nielsen {\it et al.} the properties of the state are emphasized. We shall in particular explore the relationship between structural properties of a state (i.e., entanglement) and its dynamics.

There is also a foundational motivation for this. Since the inception of quantum mechanics (QM), there has been an uneasy dichotomy between two points of view: is QM a fundamental description of nature or merely an algorithm to calculate probabilities for outcomes of experiments? The friction between these two viewpoints comes from the manifestly non-classical phenomena QM predicts, constrains or allows. This list of phenomena includes interference, the uncertainty principle, non-locality through entanglement (Bell inequalities) or otherwise \cite{Bennett:1999}, quantum teleportation and no-cloning theorems \cite{Wootters:1982}. The relationships between these phenomena remain unclear \cite{Barrett:2007ip, Oppenheim:2010, Popescu:1994, Gross:2010mc}. Our interest lies in how the structure of the allowed states of a quantum theory constrains its dynamics \cite{Barrett:2007ip, Gross:2010mc}. We suggest that the results of this paper may be extrapolated to probabilistic theories
  more general than quantum mechanics.

We utilize a symmetry of quantum states called envariance \cite{Zurek:2005pr} which emerges dynamically due to their entanglement structure. This symmetry is a consequence of the tensor product structure (TPS) of quantum states%
%
\footnote{It has been shown \cite{Barrett:2007ip} that TPSs are a generic feature of probabilistic theories with subsystems. Some form of envariance may then exist in such theories as well.}.
%
In classical deterministic theories, Cartesian products are used to define assemblies of subsystems rather than tensor products and so a symmetry equivalent to envariance does not exist. 

We shall use an operational definition of the observability of the change of a
quantum state: the trace distance (defined in Sec.~V) between the transformed
and untransformed state. It gives a measure of the average number of
measurements on different copies of the transformed state required to be able
to say with a given level of certainty that the state has changed. An upper
bound on this measure of the observability of the dynamics is derived. The
bound is only a function of the level of mixedness of the state. Another
intriguing aspect of quantum states is the mathematical equivalence of
entangled system states to mixed states representing classical ensembles of
quantum states. This connection allows the bound to apply to entangled systems
as well. 


Envariance is defined in Sec.~II. In Sec.~III, we describe how the 
information contained in the entangled subsystems of a bipartite system can 
be less than the information contained in the whole system. It is also 
shown how the mixedness of a state constrains knowledge on all non-degenerate 
observables of that state. These two qualities of quantum states are then 
used to motivate Sec.~IV where the class of invariant operations on a 
completely mixed state is derived. This symmetry is then used in Sec.~V 
to derive an upper bound on the trace distance between an untransformed 
state and a transformed one. Concluding remarks are presented in Sec.~VI.


\section{Envariance}

Envariance \cite{Zurek:2005pr,Herbut:2007qp,Paris:2005ul} is a symmetry
of entangled composite systems. We define a composite system, in
general, as a state that can be decomposed in terms of eigenstates of
two or more mutually commuting observables where subsets of the total
set of mutually commuting observables completely describe
subsystems. Therefore a particle state with quantum numbers spin and
position can be considered composite, with spin and position
describing separate subsystems.

Zurek's original use of envariance was to provide a proof of Born's
rule under `very mild' assumptions. We shall be using envariance in a
different context and thus assume Born's rule from the outset.

Consider a composite system that can be decomposed into two subsystems
$\alpha$ and $\beta$, with the state $\ket{\psi} \in \hil_\alpha
\otimes \hil_\beta$. Now suppose there exist unitary operators
$U_\alpha$ and $U_\beta$, where
%
\begin{equation}
U_\alpha \mathrel{\mathop:}= \bar{U}_\alpha \otimes \mathbb{I}_\beta,
\label{first}
\end{equation}
%
with $\mathbb{I}_\beta$ being the identity on $\hil_\beta$ and
$\bar{U}_\alpha : \tilde{\hil}_\alpha \rightarrow \tilde{\hil}_\alpha$
where $\tilde{\hil}_\alpha$ is a subspace of $\hil_\alpha$, with an analogous definition for $U_\beta$. A state $\ket{\psi}$ is said to be envariant under $U_{\alpha}$ (or $U_\beta$) if the following holds:
%
\begin{equation}
U_\alpha U_\beta \ket{\psi} = \ket{\psi}.
\label{envdef}
\end{equation}
%
We can rewrite Eq.~(\ref{envdef}) in the form
$U_{\alpha}\ket{\psi}=U_{\beta}^{\dagger}\ket{\psi}$ (Herbut's twin
unitaries \cite{Herbut:2007qp}) which will be useful later on.

Note that from now on, we shall be considering finite dimensional
Hilbert spaces only.

Suppose we have a state of the form
%
\begin{equation}
\ket{\psi} = \frac{1}{\sqrt{N}}\sum^N_{j=1} e^{i\phi_j}\ket{\alpha_j}\ket{\beta_j},
\label{third}
\end{equation}
%
where $\left\{\ket{\alpha_j}\left|j\right.\right\}$ and
$\left\{\ket{\beta_j}\left|j\right.\right\}$ form orthonormal bases
for $\tilde{\hil}_\alpha$ and $\tilde{\hil}_\beta$ respectively, $N =
\textrm{dim}(\tilde{\hil}_\alpha) = \textrm{dim}(\tilde{\hil}_\beta)$
and $\phi_j$ are arbitrary phases. These states are envariant under
{\it all} unitary transformations of $\tilde{\hil}_\alpha$ (or
$\tilde{\hil}_\beta$). When $\tilde{\hil}_\alpha = \hil_\alpha$, the
state is maximally entangled (for subsystems $\alpha$ and $\beta$) and
the group of envariant transformations is the group of all unitary
transformations of $\hil_\alpha$ alone (i.e., they can be decomposed as
in Eq.~(\ref{first})).

Consider now the case of a state with Schmidt decomposition
%
\begin{equation}
\ket{\Omega} = \sum^n_{j=1} c_j e^{i\phi_j}\ket{\alpha_j}\ket{\beta_j},
\label{fourth}
\end{equation}
%
with $c_j \in \mathbb{R}$ such that $c_i \neq c_j$ for $i \neq j$, i.e., the
coefficients of the Schmidt decomposition have unequal norms. In
this case, the group of envariant transformations includes only
relative (and overall) phase changes between the components
$\ket{\alpha_i}\ket{\beta_i}$, i.e., unitaries of the form
%
\begin{equation}
U_\alpha := \Big(\sum^n_{j=1}e^{i\lambda_j}\ket{\alpha_j}\bra{\alpha_j} + \sum^N_{j=n+1}\ket{\alpha_j}\bra{\alpha_j}\Big)\otimes\mathbb{I}_\beta,
\end{equation}
%
\begin{equation}
U_\beta := \Big(\sum^n_{k=1}e^{-i\lambda_k}\ket{\beta_k}\bra{\beta_k} + \sum^M_{k=n+1}\ket{\beta_k}\bra{\beta_k}\Big)\otimes\mathbb{I}_\alpha,
\end{equation}
%
where $\textrm{dim}(\hil_\alpha) = N, \textrm{dim}(\hil_\beta) = M$
and $\lambda_j \in (0,2\pi)$ $\forall$ $j$. These unitaries have the desirable property
%
\begin{equation}
U_\alpha U_\beta \ket{\Omega} = \ket{\Omega}.
\end{equation}

The most general case is where some of the coefficients $c_i$ are
equal and others are not. For the subspaces of $\hil_\alpha$ spanned
by the components whose coefficients are equal, we have envariance
over the entire subspace. For the rest of the space, it is only
relative phases of the components with unequal coefficients that can
be envariantly transformed.

\section{Allowed States}

The emergence of envariance is a reflection of the property of
entangled quantum states, where complete knowledge of the entire system
(i.e., the state being pure) means incomplete knowledge of the
subsystems. This can be understood in several ways:

1. The reduced density matrices, tracing out $\alpha$ or $\beta$ ($\mathrm{tr}_\alpha [\rho]$ or $\mathrm{tr}_\beta [\rho]$ for some pure $\rho$), have non-zero von Neumann entropy, leaving a mixed state partially
equivalent to a classical lack of knowledge about the subsystem. However, the number of invariant degrees of freedom is only indirectly related to the amount of entanglement as mentioned in \cite{Paris:2005ul}. We shall consider this point in more detail later on. 

2. Consider a Bell state,
%
\begin{equation}
\ket{\nu} = \tfrac{1}{\sqrt{2}}\left(\ket{\uparrow}_1\ket{\downarrow}_2 + \ket{\downarrow}_1\ket{\uparrow}_2\right),
\end{equation}
%
which is maximally entangled and as such has an SU(2) subgroup of
envariant transformations which we can parametrize by the Pauli
matrices,
%
\begin{equation}
e^{i\vec{\theta} \cdot \vec{\sigma}_1}\ket{\nu} = e^{i\vec{\theta} \cdot \vec{\sigma}_2}\ket{\nu},
\end{equation}
%
where $\sigma^i_j$ is the $i^{\textrm{th}}$ Pauli matrix for the
$j^{\textrm{th}}$ particle. Thus, rotating the spin of particle one is
the same as rotating the spin of particle two instead. This implies
only the {\it relative} orientations of the rays within the subsystem
Hilbert spaces are known. Zurek has cited a similar idea as his
motivation for using envariance \cite{Wheeler:1984dy} and calls it the
`relativity of quantum observables'. The situation can be said to have
a kind of Machianity \cite{Barbour:1994ri}, analogous to the situation where the
universe consists of point particles and only relative distances
between them are known, not global displacement or orientation. The state only contains information about the correlations between the
particles. 

3. For maximally entangled subsystems, the probabilities for fine grained (non-degenerate) measurement outcomes of a subsystem whose reduced density matrix has its maximum von Neumann entropy become equal {\it in any basis}. This can be seen with the use of envariance, which is equivalent to a basis ambiguity of the subsystems. For instance, with the Bell state the probabilities for a
particular particle to be spin up or down in the $z$-direction are the
same while the probabilities for the particle to be spin left or right
in the $x$ or $y$ direction are also the same. This is in contrast to
an unentangled spin-$1/2$ particle where there always exists a
direction where the spin is definitely known.

This last example is a special case of a phenomenon that does not
apply to classical physics. The uncertainty principle is usually
applied to pure states but the situation changes for mixed states,
such that the bounds on the uncertainties for incompatible observables
become more strict. To show this, we utilize the concavity of the
expression $-x\textrm{ln}x$ (with $x \in \mathbb{R}^+$) \cite{Wehrl:1978}
such that for a density matrix $\rho$ and a fine grained basis
$\left\{\ket{i}\right\}$, $\bra{i}j\rangle =
\delta_{ij}$, $\sum_i \ket{i}\bra{i} = \mathbb{I}$, the von Neumann
entropy $S(\rho)$ has the property:
%
\begin{align}
S(\rho) :&= -\mathrm{tr}[\rho\textrm{ln}\rho] \nonumber \\
&= \sum_i \bra{i}-\rho\textrm{ln}\rho\ket{i} \nonumber \\
&\leq -\sum_i \bra{i}\rho\ket{i}\textrm{ln}\bra{i}\rho\ket{i}.
\end{align}

Choosing $\left\{\ket{i}\left|i\right.\right\}$ to be the eigenstates
of a fine grained observable, then $\bra{i}\rho\ket{i}$ is the
probability to measure outcome ``$i$'' such that the Shannon entropy
of said observable (call it $O$) is given by
\begin{equation}
H_O(\rho) = -\sum_i \bra{i}\rho\ket{i}\textrm{ln}\bra{i}\rho\ket{i}.
\end{equation}
%
Thus,
\begin{equation}
S(\rho) \leq H_O(\rho).
\end{equation}
%
For an alternative proof see \cite{Schumacher:1996}.

This applies to all fine grained observables of the system described
by $\rho$. In the case of the Bell state, the reduced density matrix
obtained by tracing out one of the particles has maximum von Neumann
entropy such that all non-trivial observables of the subsystem also
have maximum Shannon entropy. Thus if a subsystem contains
correlations with another, the information we have about the subsystem
is more constrained than in the case of classical physics where
Shannon entropies of `incompatible' observables are allowed to be
independent. Moreover, the entanglement may be genuine or due to
purification of a mixed state.

\section{Non-Unitary Symmetry and Observable Dynamics}

Intuitively, when one lacks knowledge of a system, one expects our
ability to distinguish the dynamics of the system to be lessened. We
have seen that in the case of mixed quantum states, our knowledge of
the system is less than allowed classically. For cases of completely
mixed states, envariance can be used to give us some interesting
alternative insights into this loss of information. The consideration
of mixed density matrices, however, allows us to extend envariance to
non-unitary operations. The use of such symmetries will then allow us
to examine restrictions on our abilities to distinguish the dynamics
of mixed states.

Consider again the state $\ket{\psi}$ of Eq.~(\ref{third}) in the case where subsystem $\alpha$ is maximally entangled with subsystem $\beta$. Suppose we
have a linear operation $\bar{\mu}_\alpha:\hil_\alpha \rightarrow
\hil_\alpha$ such that $\mu_\alpha := \bar{\mu}_\alpha \otimes
\mathbb{I}_\beta$. We wish to find a similarly defined $\mu_\beta$
such that $\mu_\alpha\ket{\psi} =
\mu_\beta\ket{\psi}$. Note that we do not assume $\mu_\alpha$ to be unitary.
%
\begin{equation}
\mu_\alpha \ket{\psi} = \frac{1}{\sqrt{N}}\sum^N_{j=1}e^{i\phi_j}\left(\bar{\mu}_\alpha\ket{\alpha_j}\right)\ket{\beta_j}.
\end{equation}
%
Define $\bra{\alpha_i}\bar{\mu}_\alpha\ket{\alpha_j} := \mu_{ij}$ such that
%
\begin{align}
\mu_\alpha\ket{\psi} &= \frac{1}{\sqrt{N}}\sum^N_{j=1}e^{i\phi_j}\Big(\sum^N_{i=1}\mu_{ij}\ket{\alpha_i}\Big)\ket{\beta_j} \nonumber \\
&= \frac{1}{\sqrt{N}}\sum^N_{i=1}\ket{\alpha_i}\Big(\sum^N_{j=1}e^{i\phi_j}\mu_{ij}\ket{\beta_j}\Big) \nonumber \\
&= \frac{1}{\sqrt{N}}\sum^N_{i=1}\ket{\alpha_i}\ket{\tilde{\beta}_i},
\end{align}
%
where $\ket{\tilde{\beta}_i} := \sum^N_{j=1}e^{i\phi_j}\mu_{ij}\ket{\beta_j}$. For non-unitary $\mu_\alpha$, $\{\ket{\tilde{\beta}_j}\}$ need not be orthonormal.

Define
%
\begin{align}
\mu_\beta:=\mathbb{I}_\alpha&\otimes\sum^N_{k=1}e^{i\phi_k}\ket{\tilde\beta_k}\bra{\beta_k}:=\mathbb{I}_\alpha\otimes\bar\mu_\beta, \\
\therefore\mu_\alpha\ket{\psi}&=\mu_\beta\ket{\psi}.
\end{align}
%
Let's now consider the completely mixed density matrix for $\alpha$,
%
\begin{equation}
\rho_\alpha := \mathrm{tr}_\beta [\ket{\psi}\bra{\psi}] = \frac{1}{N}\mathbb{I}_\alpha.
\end{equation}
%
Suppose we have a quantum operation $\mathcal{E}_\alpha(\rho)=\rho'$ that is given in an operator-sum representation \cite{Nielsen:2004};
%
\begin{equation}
\mathcal{E}_\alpha(\rho) = \sum^K_{k=1}E_{k\alpha}\rho E^{\dagger}_{k\alpha},
\end{equation}
%
where $E_{k\alpha}$ are linear maps $E_{k\alpha} : \hil_\alpha
\rightarrow \hil_\alpha$. Also suppose $\mathcal{E}_\alpha$ is trace preserving,
such that $\sum^K_{k=1} E^{\dagger}_{k\alpha}E_{k\alpha} =
\mathbb{I}_\alpha$. The effect of $\mathcal{E}_\alpha$ upon $\rho_\alpha$ is
then
%
\begin{align}
\mathcal{E}_\alpha(\rho_\alpha) &= \sum^K_{k=1} E_{k\alpha}\mathrm{tr}_\beta\left[\ket{\psi}\bra{\psi}\right]E^{\dagger}_{k\alpha} \nonumber \\
&= \mathrm{tr}_\beta\Big[\sum^K_{k=1}E_{k\alpha}\ket{\psi}\bra{\psi}E^{\dagger}_{k\alpha}\Big].
\end{align}
%
Since $E_{k\alpha}$ is of the form $\bar{\mu}_\alpha$, there exists an
$E_{k\beta}$ such that $E_{k\alpha}\ket{\psi} = E_{k\beta}\ket{\psi}$
and similarly $\bra{\psi}E^\dagger_{k\alpha} = \bra{\psi}E^\dagger_{k\beta}$. Then
%
\begin{align}
\mathcal{E}_\alpha(\rho_\alpha) &= \mathrm{tr}_\beta\Big[\sum^K_{k=1}E_{k\beta}\ket{\psi}\bra{\psi}E^\dagger_{k\beta}\Big] \nonumber \\
&= \mathrm{tr}_\beta\Big[\sum^K_{k=1}E^\dagger_{k\beta}E_{k\beta}\ket{\psi}\bra{\psi}\Big].
\end{align}
%
If $\sum^K_{k=1}E^\dagger_{k\beta}E_{k\beta} = \mathbb{I}_\beta$, then
%
\begin{equation}
\mathcal{E}_\alpha(\rho_\alpha) = \rho_\alpha.
\end{equation}
%
The completely mixed state is then invariant under $\mathcal{E}_\alpha$
if $\mathcal{E}_\alpha$ is trace preserving and satisfies the condition $\sum^K_{k=1}E^{\dagger}_{k\beta}E_{k\beta} = \mathbb{I}_\beta$, which is equivalent to the condition
%
\begin{equation}
\sum^K_{k=1}E_{k\alpha}E^{\dagger}_{k\alpha} = \mathbb{I}_\alpha.
\label{Condition}
\end{equation}

Thus, all dynamics of this form, acting on the state $\rho_\alpha$, are completely unobservable. Operations satisfying these conditions include unitary $\mathcal{E}(\rho) = U\rho U^\dagger$, perfect premeasurements $\mathcal{E}(\rho) = \sum^K_{k=1}P_k\rho P_k$ where $P_k$ are projectors of a complete basis and combinations of unitary and premeasurement transformations. Interestingly, generalized measurements \cite{Nielsen:2004} where the outcome is unknown do not necessarily satisfy the condition (\ref{Condition}).

\section{Upper Bound}

Let us now consider a general bipartite state with Schmidt
decomposition of the form previously given in Eq.~(\ref{fourth}), $\ket{\Omega} = \sum^n_{j=1}c_je^{i\phi_j}\ket{\alpha_j}\ket{\beta_j}$ except the
coefficients $c_j$ are arbitrary up to normalization of $\Omega$. If
the coefficients are not all equal and/or $\tilde{\hil}_\alpha <
\hil_\alpha$, then the group of envariant operations is greatly
reduced. The set of all operations that leave
$\mathrm{tr}_\beta\left[\ket{\Omega}\bra{\Omega}\right]$ invariant is
also reduced. This limits the previous proof of the unobservability of
the dynamics of $\alpha$ for cases where the state is not completely
mixed. However, a large reduction in the symmetry can occur with only
a very small reduction in the von Neumann entropy of
$\alpha$ \cite{Paris:2005ul}. This suggests that even with a large
reduction in the set of symmetries, the full set may apply in a
partial sense. Mixed states with less than maximum von Neumann
entropy may still have some form of limitations on their observable
dynamics for the full set of trace preserving operations satisfying
Eq.~(\ref{Condition}). This turns out to be the case.

To see this, we initially rewrite $\Omega$. Let us extend the sum\footnote{However, the choice of the size of the extension may be chosen to be smaller depending on whether $\mathcal{E}$ leaves certain subspaces of $\hil_\alpha$ invariant.} over $j$ from one to $N$ and define $c_j=0$ for $n + 1 \leq j \leq N$. For $M < N$, we enlarge $\hil_\beta$ until the dimensionalities are equal. We then decompose $\Omega$ into two parts, one that is maximally symmetric over $\hil_\alpha$ and the rest of the state;
%
\begin{equation}
c_j = \frac{1}{\sqrt{N}} + d_j
\end{equation}
%
where $d_j := c_j - 1/\sqrt{N}$, such that
%
\begin{equation}
\ket{\Omega} = \frac{1}{\sqrt{N}} \sum^N_{j=1} e^{i\phi_j}\ket{\alpha_j}\ket{\beta_j} + \sum^N_{j=1}d_je^{i\phi_j}\ket{\alpha_j}\ket{\beta_j}.
\end{equation}
%
Define
%
\begin{align}
\ket{\Omega_1} &= \frac{1}{\sqrt{N}}\sum^N_{j=1}e^{i\phi_j}\ket{\alpha_j}\ket{\beta_j} \\
Q\ket{\Omega_2} &= \sum^N_{j=1}d_j e^{i\phi_j}\ket{\alpha_j}\ket{\beta_j},
\end{align}
%
where $Q$ is a constant to normalize $\Omega_2$ such that $Q =
\sqrt{\sum_j d^2_j}$. $Q$ will be our measure of the difference between
the state $\Omega$ and the maximal symmetric state $\Omega_1$. $Q^2$
is just the $\chi^2$ value between the distribution of amplitudes
$\left\{c_j\right\}$ and the constant distribution
$1/\sqrt{N}$.

We now utilize a measure of the distinguishability of quantum states,
the trace distance, defined as
%
\begin{equation}
D(\rho,\sigma):=\tfrac{1}{2}\mathrm{tr}\vert\rho-\sigma\vert,
\end{equation}
%
where $\rho$ and $\sigma$ are density matrices and $\vert
X\vert:=\sqrt{X^\dagger X}$ is the positive square root of $X^\dagger
X$ (defined by taking a spectral decomposition $X^\dagger X =
\sum_ie_i\ket{x_i}\bra{x_i}$ and taking the positive square roots of
the eigenvalues $\sqrt{X^\dagger X} = \sum_i\sqrt{e_i}\ket{x_i}\bra{x_i}$). 

It can be shown that \cite{Nielsen:2004}
%
\begin{equation}
D(\rho,\sigma)=\stackrel{max}{P}\mathrm{tr}[P(\rho-\sigma)],
\end{equation}
%
where $P$ is a projector and the maximization is taken over all
possible projectors. This gives a clear physical interpretation of the
trace distance. If an experimentalist wanted to distinguish whether
they had the state $\rho$ or $\sigma$, the trace distance gives the
maximum possible difference in probabilities for a projective
measurement outcome for the two states. Thus the trace distance
measures how easy it is, on average, to tell two states apart. For
instance, if for two states $D=1$, it is in principle possible to do a
projective measurement where the probability of getting a confirmatory
result for one state is one while the other is zero and hence only one
measurement is ever needed to distinguish the states.

Suppose we prepare the state $\rho = \ket{\Omega}\bra{\Omega}$ and it
is acted on by $\mathcal{E}_\alpha$, such that
$\sum^K_{k=1}E^\dagger_{k\alpha}E_{k\alpha} = \mathbb{I}_\alpha =
\sum^K_{k=1}E_{k\alpha}E^\dagger_{k\alpha}$. In the case where $Q =
  0$, $\ket{\Omega} = \ket{\Omega_1}$ and
%
\begin{equation}
\mathcal{E}_\alpha\left(\rho_{Q=0}\right) = \mathcal{E}_\beta\left(\rho_{Q=0}\right).
\end{equation}
%
As these two states are equal, they are indistinguishable. For the
general case where $Q$ may not be zero, a measure for the
distinguishability of the two states can be given by the trace
distance,
%
\begin{equation}
D_{\alpha\beta} := D\Big(\mathcal{E}_\alpha(\rho),\mathcal{E}_\beta(\rho)\Big).
\end{equation}
%
We have computed an upper bound on $D_{\alpha\beta}$ (see Appendix),
%
\begin{equation}
D_{\alpha\beta} \leq 2\sqrt{1-\left| 1-Q^{2}+\tfrac{1}{4}Q^{4}\right|}.
\label{upperbound}
\end{equation}

For the case $Q=0$, the bound also shows indistinguishability of the
two states. For small $Q$ however, the ability of an experimentalist
to tell whether $(\mathcal{E}_\alpha,\rho)$ or $(\mathcal{E}_\beta,\rho)$
happened is still bound at least up to $Q=\sqrt{2-\sqrt{3}} \approx
0.5$. For cases where $\hil_\beta$ has to be enlarged and $\beta$ is
considered a real subsystem, $\mathcal{E}_\beta$ may not strictly be
physical.

If $\beta$ is an ancilla subsystem used to purify $\alpha$ or the
experimentalist does not have access to subsystem $\beta$, then we can
ask about our ability to tell whether $\mathcal{E}_\alpha$ has happened
at all. This can be quantified by;
%
\begin{align}
D_{\alpha} :&= D\Big(\mathrm{tr}_\beta[\mathcal{E}_\alpha(\rho)],\mathrm{tr}_\beta[\mathcal{E}_\beta(\rho)]\Big) \nonumber \\
&= D\Big(\mathcal{E}_\alpha(\mathrm{tr}_\beta[\rho]),\mathrm{tr}_\beta[\rho]\Big).
\end{align}
%
Because the partial trace over $\beta$ is trace preserving, it is
also bound,
%
\begin{equation}
D_\alpha\leq D_{\alpha\beta}\leq2\sqrt{1-\left|1-Q^2+\tfrac{1}{4}Q^4\right|}.
\label{thirty-three}
\end{equation}
%
This is our main result, which shows that our original intuition has some legitimacy.

Contrast this with cases where $Q$ is too large such that
$D_\alpha$ is unbound. Take a pure state $\sigma =
\ket{\alpha_m}\bra{\alpha_m}$, for example, which has a mixedness of $Q = \big(2 - 2/\sqrt{N}\big)^\frac{1}{2} < \sqrt{2 - \sqrt{3}}$ for $N \geq 2$ and a unitary transformation of the form
%
\begin{equation}
\mu_\alpha=\sum_k\ket{\tilde\alpha_k}\bra{\alpha_k},
\end{equation}
%
such that $\bra{\tilde\alpha_m}\alpha_m\rangle=0$. Then one can see
that the transformed and untransformed states have zero overlap;
%
\begin{equation}
\bra{\alpha_m}\mu_\alpha\sigma\mu^{\dagger}_\alpha\ket{\alpha_m} = 0,
\end{equation}
%
or
%
\begin{equation}
D(\mu_\alpha\sigma\mu^\dagger_\alpha,\sigma) = 1,
\end{equation}
%
making the two states in principle easily distinguishable.

Suppose $\alpha$ experiences a perfect premeasurement such that
%
\begin{equation}
\sigma \rightarrow \sum^N_{i=1}P_i\sigma P_i,
\end{equation}
%
where $P_i = \ket{A_i}\bra{A_i}, \left\{\ket{A_i}\right\}$
forms a complete orthonormal basis for $\hil_\alpha$ and
$\ket{\alpha_m} = \sum^N_{i=n}(1/\sqrt{N})\ket{A_i}$. The fidelity
(defined in Appendix) can be calculated;
%
\begin{equation}
F(\sigma,\sum_iP_i\sigma P_i)^2 = \bra{\alpha_m}\sum_iP_i\sigma P_i\ket{\alpha_m} = \frac{1}{N}.
\end{equation}
%
In this case, the fidelity bounds the trace distance
%
\begin{equation}
1 - F(\sigma,\sum_iP_i\sigma P_i)^2 = 1 - \frac{1}{N} \leq D(\sigma,\sum_iP_i\sigma P_i).
\end{equation}
%
Thus, the distinguishability of the premeasurement
process on the pure state $\sigma$ becomes progressively constrained to be large as $N$ increases.

Finally, we note that the bound may be extended to mixed states of a
composite $\alpha$ and $\beta$ system. Let
%
\begin{equation}
\bar{\rho} = \sum_m r_m\rho_m,
\end{equation}
%
where $\sum_mr_m = 1$ and $\rho_m$ is a pure density matrix of the
composite system $\forall m$. Define $Q_m$ as the $Q$ measure of the
mixedness of the $\mathrm{tr}_\beta(\rho_m)$ state and define
$\mathcal{E}_\alpha$ and $\mathcal{E}_\beta$ in the usual way. Then, using
the convexity of the trace distance,
%
\begin{align}
D\Big(\mathcal{E}_\alpha(\bar{\rho}),\mathcal{E}_\beta(\bar{\rho})\Big) &\leq \sum_m r_m D\Big(\mathcal{E}_\alpha(\rho_m), \mathcal{E}_\beta(\rho_m)\Big) \nonumber \\
&\leq 2\sum_m r_m \sqrt{1 - \left|1 - Q^2_m + \tfrac{1}{4}Q^4_m\right|}.
\end{align}

Thus, the distinguishability of the dynamics of $\alpha$ in a mixed
composite state is bound by the average of the bounds of the pure
states $\rho_m$.

\section{Remarks}

Taking envariance as a starting point, we have shown the existence of a large
class of symmetries of completely mixed states that include
non-unitary operations. These symmetries were then used to derive an
upper bound on the dynamics of general mixed states subject to the
condition (\ref{Condition}). Moreover, the bound involves only
the level of mixedness of the state, quantified by our measure $Q$.

The meaning of the condition (\ref{Condition}) is unclear at this stage. We suggested that the bound (\ref{thirty-three}) was related to the von Neumann entropy being a lower bound on the Shannon entropies of the fine grained observables of the state. Could trace preservation and Eq.~(\ref{Condition}) then be related to changes in von Neumann entropy? For example, both unitary and perfect premeasurement operations which satisfy Eq.~(\ref{Condition}) cannot decrease the von Neumann entropy. In contrast, generalized measurements which don't satisfy Eq.~(\ref{Condition}) can. To see this, take for example measurement operators of a two level system $M_1 = \ket{0}\bra{0}$ and $M_2 = \ket{0}\bra{1}$. The left hand side of Eq.~(\ref{Condition}) with $E_{i\alpha} = M_i$ does not equal unity; $M_1M^\dagger_1 + M_2M^\dagger_2 \neq \mathbb{I}$. We speculate that there may be a connection between the conditions of trace preservation and Eq.~(\ref{Condition}) and an inability to decrease the von Neumann entropy.

\begin{acknowledgments}
I thank R.J. Crewther and Lewis C. Tunstall for a critical reading of the manuscript.  I acknowledge technical support from S. Underwood, Dale S. Roberts, and A. Casey.  This work is supported by the Australian Research Council.
\end{acknowledgments}

\setcounter{secnumdepth}{-1}
\appendix
\section{Appendix: Proof of upper bound of trace distance}
\label{app1}
To prove Eq.~(\ref{upperbound}), we shall need a few identities. Using
the definition of $d_i$ and taking $\Omega$ to be normalized, we find;
\begin{align}
\sum^N_{i=1}c^2_i &=1=\sum^N_{i=1}\left(\frac{1}{\sqrt{N}}+d_i\right)^2 \nonumber \\
&=1+\sum^N_{i=1}d_i^2+\frac{2}{\sqrt{N}}\sum_{i=1}^Nd_i \nonumber \\
\therefore \sum_{i=1}^Nd_i^2 &=-\frac{2}{\sqrt{N}}\sum^N_{i=1}d_i
\label{coefficients}
\end{align}
We also compute the overlap
\begin{equation}
\bra{\Omega_1}\Omega_2\rangle=\frac{1}{Q\sqrt{N}}\sum_{i=1}^Nd_i=-\frac{1}{2}Q,
\end{equation}
using Eq.~(\ref{coefficients}). Defining
\begin{align}
\rho_{11}&:=\ket{\Omega_1}\bra{\Omega_1} \nonumber \\
\rho_{12}&:=\ket{\Omega_1}\bra{\Omega_2} \nonumber \\
\rho_{21}&:=\ket{\Omega_2}\bra{\Omega_1} \nonumber \\
\rho_{22}&:=\ket{\Omega_2}\bra{\Omega_2},
\end{align}
we have
\begin{align}
\rho &= \ket{\Omega}\bra{\Omega} \nonumber \\
&= \rho_{11}+Q(\rho_{12}+\rho_{21})+Q^2\rho_{22}.
\end{align}
The $\rho$s have some useful relationships, namely,
\begin{align}
\rho_{11}^2 &= \rho_{11}, \rho_{22}^2 = \rho_{22}, \rho_{12}\rho_{21} = \rho_{11}, \nonumber \\
\rho_{12}\rho_{12} &= -\tfrac{Q}{2}\rho_{12}, \rho_{21}\rho_{12} = \rho_{22}, \rho_{21}\rho_{21} = -\tfrac{Q}{2}\rho_{21}.
\end{align}
The trace distance has particular properties which shall also be
used. Thus,
\begin{align}
D_{\alpha\beta}:&=D\Big(\mathcal{E}_\alpha(\rho),\mathcal{E}_\beta(\rho)\Big) \nonumber \\
&\leq D\Big(\mathcal{E}_\alpha(\rho),\mathcal{E}_\alpha(\rho_{11})\Big)+D\Big(\mathcal{E}_\alpha(\rho_{11}),\mathcal{E}_\beta(\rho)\Big) \nonumber \\
&=D\Big(\mathcal{E}_\alpha(\rho),\mathcal{E}_\alpha(\rho_{11})\Big)+D\Big(\mathcal{E}_\beta(\rho_{11}),\mathcal{E}_\beta(\rho)\Big) \nonumber \\
&\leq D(\rho,\rho_{11})+D(\rho_{11},\rho) \nonumber \\
&=2D(\rho_{11},\rho).
\end{align}
Because $\rho$ and $\rho_{11}$ are pure,
\begin{equation}
D(\rho_{11},\rho)=\sqrt{1-F(\rho_{11},\rho)^2},
\end{equation}
where $F$ is the fidelity;
\begin{align}
F(\rho_{11},\rho) :&= \mathrm{tr}\Big[\big(\rho_{11}^{1/2}\rho\rho_{11}^{1/2}\big)^{1/2}\Big] \nonumber \\
&=\mathrm{tr}\Big[\big(\rho_{11}\rho\rho_{11}\big)^{1/2}\Big] \nonumber \\
&=\mathrm{tr}\Big[\big(\rho_{11}[\rho_{11}+Q(\rho_{12}+\rho_{21})+Q^2\rho_{22}]\rho_{11}\big)^{1/2}\Big] \nonumber \\
&=\mathrm{tr}\Big[\big(\rho_{11}-\tfrac{1}{2}Q^2\rho_{11}-\tfrac{1}{2}Q^2\rho_{11}+\tfrac{1}{4}Q^4\rho_{11}\big)^{1/2}\Big] \nonumber \\
&=\sqrt{1-Q^2+\tfrac{1}{4}Q^4}.
\end{align}
Thus,
\begin{equation}
D_{\alpha\beta}\leq 2\sqrt{1-\left|1-Q^2+\tfrac{1}{4}Q^4\right|}.
\end{equation}

\vspace{20pt}

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\end{document}
